Pulling will require less work.
The force required to pull is
F = 50a / (sqrt(3) + .2)
where a is the acceleration due to gravity, or a little more than 25a Newtons.
The force required to push is
F = 50a / (sqrt(3) - .2)
or about 37a Newtons.
In both cases, the horizontal component of the force, which would be used to calculate the work done, is the force multiplied by
sqrt(3) / 2.
The work lost to friction is the horizontal force times 15 meters.
To compute the force required to pull and the force required to push, I decomposed the force into its horizontal and vertical components, using sine (30 degrees) = 1/2 and cosine (30 degrees) = sqrt(3) / 2. The vertical component of the force either adds to (pushing) or subtracts from (pulling) the "weight" of the 50kg mass, or 50a, giving the "normal" force. The friction, which must be equal to the horizontal component of the force, is .2 times the normal force.
2006-07-08 14:01:34
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answer #1
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answered by ? 6
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A push or pull is a force. Force has magnitude and direction.
How will you represent a push or pull in a figure?
A 10 newton force acts on an object at angle 30 degree to the horizontal.
To represent the force in the figure, assume that the object is at the origin.
We will draw a line of 5 cm (scale 5cm = 10 newton) from the origin at angle of 30 degree to the x axis. We will draw a small arrow at the end of this line.
Now seeing the figure can you say whether the force is a ‘PULL’ or ‘PUSH’?
Whether it is a pull or push, the force is having one magnitude and one direction.
In the given problem since the force is applied at angle of 30 degree to the horizontal whether it is a pull or push both will have the same effect as here the force is the same.
In certain problems, push and pull will mean different directions of force.
For example, suppose we are PULLING (we are in front of the box) a box along a horizontal floor, by a force at an angle of 45 degree to the horizontal. The same box is PUSHED (we are at the back of the box) along the horizontal, again at an angle 45 degree. What is the difference in these two cases?
We have committed a mistake in measuring the angle of the forces in the latter case. A careful analysis will show that the angle in the latter case is minus 45 degree ( below the horizontal) and not 45 degree. Therefore, in this case the push is not equal to the pull.
If we apply a force of push at exactly 45 degree, (from the bottom of the box pointing upward at angle of 45 degree) then the push and pull will be the same.
Therefore if the angle of the force (direction) is not changed, push and pull is the same.
2006-07-08 15:52:42
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answer #2
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answered by Pearlsawme 7
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Peter T is correct. Assuming the crate is not lifted by the pulling action, then no work is done in the vertical direction. Kinetic friction is directly proportional to the perpendicular force between the two surfaces (the crate and the supporting surface). Pushing adds a downward component, thus increasing the force and the work required to overcome friction. Pulling at 30 degrees provides an upward component that reduces this force; and, consequently, the work (energy) lost to friction.
2006-07-08 13:51:30
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answer #3
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answered by oscarsnerd 2
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Less work.
Sin 30^o = 0.5
(For convenience, g = 10 m/s^2)
If the student pulls, 50% of his effort goes into lifting the crate, making it that much lighter, and therefore the 500 N weight of the crate will decrease. If he pushes, then 50 % of the force will be added to the 500 N weight.
Was the Law of Conservation of Energy broken? No, since one needs to add the extra thermal energy generated due to greater friction in the case of pushing.
2006-07-08 13:36:23
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answer #4
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answered by flandargo 5
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According to the laws of conservation of energy, the student will do the same amount of work in any of these scenarios.
However.
The work will only be considered part of the chain of events if it is actually successful, and some approaches may not work at all if the student's arms are not strong enough (for instance) to pull, but her legs are strong enough that she can put her back against the crate and use her legs to push the crate along.
Hope that clarifies rather than obfuscates!
2006-07-08 13:23:18
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answer #5
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answered by Anonymous
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Pulling is less work.
Draw a picture.
Pushing at 30 degrees means part of the student's applied force is into the ground, thus increasing the friction.
Pulling at 30 degrees means part of the student's applied force is out of the ground, thus decreasing friction.
2006-07-08 13:22:42
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answer #6
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answered by wdmc 4
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If the student were to push and pull at 180o the result is the same. But at 30o, pulling will be easier. The vertical component from pushing is downwards and from pulling is upwards.
2006-07-08 13:39:41
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answer #7
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answered by Peter T 2
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If you can get the same force in either direction than pushing or pulling won't matter. It is what 'the student' is better at
2006-07-08 13:22:46
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answer #8
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answered by satanorsanta 3
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the less work for pulling people are correct
and some of their discussion is helpful
but to understand this you need to calculate the normal force (that is force perpendicular to the frictional resistance)
the normal force times the frictional coeficient tells you the resistance from friction
with part of the force you add going down, the normal force is higher, and the frictional resistance is higher, opposite if part of the force is away from the normal force
2006-07-08 14:10:32
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answer #9
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answered by enginerd 6
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The work will be the same.
2006-07-08 13:20:59
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answer #10
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answered by Engineer 6
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